Question

If the line $y=mx$ bisects the area enclosed by the lines $x=0,$ $y=0,$ $x=\frac{3}{2}$ and the curve $y=1+4x-x^2,$ then $12m$ is equal to

Answer 1

Area $OABC$
$\underset{0}{\overset{3/2}{\int }}(1+4x-x^2)dx={[x+2x^2-\frac{x^3}{3}]}_0^{3/2}=\frac{39}{8}$
If $y=mx$ passes through $B$ then
$\text{Area of OAB}=\frac{1}{2}\cdot \frac{3}{2}\cdot \frac{19}{4}=\frac{57}{16}$
Which means for bisection it should pass through a point below $B,$
Let it be $P,$ so for bisection $\text{ar}\left(△OAP\right)=\frac{39}{8}\cdot \frac{1}{2}$
$\Rightarrow \frac{1}{2}\cdot \frac{3}{2}\cdot \frac{3}{2}m=\frac{39}{8}\cdot \frac{1}{2}$
$\Rightarrow m=\frac{39}{8}\cdot \frac{4}{9}=\frac{26}{12}$
$\Rightarrow 12m=26$